Answer:
48.75 g of AgCl
11.60 g of H₂S
Solution:
The Balance Chemical Equation is as follow,
Ag₂S + HCl → AgCl + H₂S
<u>Calculate amount of AgCl produced</u><u>,</u>
According to equation,
247.8 g (1 mol) of Ag₂S produces = 143.32 g (1 mol) of AgCl
So,
84.3 g of Ag₂S will produce = X g of AgCl
Solving for X,
X = (84.3 g × 143.32 g) ÷ 247.8 g
X = 48.75 g of AgCl
<u>Calculate amount of H</u><u>₂</u><u>S produced</u><u>,</u>
According to equation,
247.8 g (1 mol) of Ag₂S produces = 34.1 g (1 mol) of H₂S
So,
84.3 g of Ag₂S will produce = X g of H₂S
Solving for X,
X = (84.3 g × 34.1 g) ÷ 247.8 g
X = 11.60 g of H₂S
Answer:
Combustion is the reaction type if you meant to put a "=" in between C3H2 and H2O
The mass in grams of NH₃ produced from the reaction is 3.4 g
<h3>Balanced equation</h3>
We'll begin by writing the balanced equation for the reaction. This illustrated below:
N₂ + 3H₂ -> 2NH₃
From the balanced equation above,
1 dm³ of N₂ reacted to produced 2 dm³ NH₃
<h3>How to determine the volume of NH₃ produced</h3>
From the balanced equation above,
1 dm³ of N₂ reacted to produced 2 dm³ NH₃
Therefore,
2.24 dm³ of N₂ will react to produce = 2.24 × 2 = 4.48 dm³ of NH₃
<h3>How to determine the mass of NH₃ produced</h3>
We'll begin by obtained the mole of 4.48 dm³ of NH₃. Details below:
22.4 dm³ = 1 mole NH₃
Therefore,
4.48 dm³ = 4.48 / 22.4
4.48 dm³ = 0.2 mole of NH₃
Finally, we shall determine the mass of NH₃ as follow:
- Molar mass of NH₃ = 17 g/mol
- Mole of NH₃ = 0.2 mole
- Mass of NH₃ =?
Mass = mole × molar mass
Mass of NH₃ = 0.2 × 17
Mass of NH₃ = 3.4 g
Learn more about stoichiometry:
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<span> Pb(NO3)2 + 2 NaBr ------> PbBr2 + 2 NaNO3
2 mol 2 mol
M(NaBr)= 102 g/mol
M(NaNO3) = 85 g/mol
1) 244g NaNO3 *1 mol NaNO3/85 g NaNO3 = 244/85 mol NaNO3
2)</span> Pb(NO3)2 + 2 NaBr ------> PbBr2 + 2 NaNO3
2 mol 2 mol<span>
x mol </span>244/85 mol
<span>
x=(2 mol*</span> 244/85 mol )/2 mol = 244/85 mol NaBr
<span>
3) </span> 244/85 mol NaBr*102g NaBr/1 mol = (244*102/85) g NaBr =292.8 g NaBr<span>
</span>
There are 11 electrons in a atom of elemental sodium.
Hope this helps!
